TMJC Chp8 Applications of Differentiation Assignment Solutions

Chapter 8 Applications of Differentiation TMJC 2024 H2 Mathematics (9758) Chapter 8 Applications of Differentiation Assignment Suggested Solutions 1 2019/SAJC Promo/Q8 The curve C has parametric equations   23 for 03 , ln , 3.t y t tx      (i) Sketch the graph of C, giving the coordinates of its endpoint(s) and the point(s) where C meets the axes. State also the equation of the vertical asymptote. [3] (ii) Find the equation of the tangent to the curve C at the point   3 2 , ln3 p p        , simplifying your answer. [5] (iii) Hence find the exact coordinates of the points Q and R where the tangent to the curve C when et  meets the x-axis and y-axis respectively. [3] (iv) Find the area of triangle OQR in exact form. [2] Q1 Solution (i) As 0, , 0.yt x   Hence vertical asymptote is 0.x  Intercept: 11 30, t xy    x O y (9, [ln(3)]ଶ) ൬1 3 , 0൰ ݔ= 0 Remember to press p and set Tmin = 0 Tmax = 3 For sketch, need to label the endpoint as a closed circle, since endpoint is at t = 3 which is included. Question asks to state equation of vertical asymptote. Keep this in mind and make sure you have a vertical asymptote in your graph. As 0t  , 3 03x t  ln t ,   2 lny t  Asymptote: 0x

Chapter 8 Applications of Differentiation TMJC 2024 (ii) 2 2d 3 d 2ln ,d d 3 x t y t tt t t   2 3 d d 1 dd d d 2 2l ln 1 n y y xx t t t t t t t      At ,t p 3 2lnd .dx py p Equation of the tangent of C at point p :       3 2 3 2 3 2 3 2lnln 3 2 2ln ln ln 3 2 2ln ln ln 3 p py p x p y p p x p p y p x p pp                     ALERT:  2ln 2lnp p  (iii) Equation of the tangent of C at point t = p :   2 3 2 2ln ln ln 3 y p x p p p       At ep ,   2 3 3 2 2ln e ln e ln ee 3 2 1 e 3 y x y x         When the tangent cuts the x – axis, 3e0 . 6y x   When the tangent cuts the axis at y – axis, 1.0 3yx   The coordinates of Q are 3e .6 , 0    The coordinates of R are 10, .3      Note: t is the parameter, p is a constant that gives a specific point So, you have to differentiate with respect to t, then substitute t = p to find the value of the gradient at that point Keyword: ‘Hence’ Means you need to use the previous result (equation of tangent at point t = p) to solve Reminder: Question wants exact coordinates so remember to answer the question Recall: Equation of line: ݕ−ݕ଴ =݉(ݔ−ݔ଴) where m is the gradient and (ݔ଴,ݕ଴) is a point on the line You need to find the value of the gradient at point P by substituting t p Note: e is a constant Use equation of tangent found in (ii), substituting ep . Simplify using ln e 1

Chapter 8 Applications of Differentiation TMJC 2024 (iv) 3 3 2 1 e 1Area of triangle 2 6 3 e units36 OQR            2 2019/JPJC Promo/Q4 Steel sheets of negligible thickness are used to make cans in the shape of a right -circular cylinder of radius r cm and height h cm. Given that a can is to hold 300 cm3 of liquid, (i) show that the external surface area, A cm2, of a closed can is 2 6002A r r  . [2] (ii) Find, using differentiation, the exact value of r which produces a can with a minimum value of A. Hence, find the minimum value of A. [4] (iii) Deduce that 2h r  when A is a minimum. [2] (iv) Liquid is dispensed into an empty can at a rate of 100 cm3/s. Find the rate of increase of the depth of the liquid, H, given that A is a minimum. [4] Reminder: Area is always positive Triangle OQR is a right angled triangle as Q and R are on the x- and y-axes respectively

Chapter 8 Applications of Differentiation TMJC 2024 2 2019/JPJC Promo/Q4 (i) Given that 2 300V r h  , 2 300h r 2 2 2 2 2 2 3002 2 6002 A r rh r r r r r                 (ii) 2 d 6004d A rr r   When d 0d A r  , 2 6004 0r r   3 150r  r 3 150         3 150  3 150         d d A r  A is minimum when 3 150r  . Alternatively: Using GC, 3 2 2 150 d 37.7 0d r A r     A is minimum when 3 150r  . Minimum A 2 3 3 150 6002 150 248 (to 3 s.f.)            (iii) A is minimum when 3 150r  . For second derivative test: Write down the value of 2 2 d 37.7d A r  at 3 150r  before comparing with 0. Question does not require A to be exact. Remember to prove that value of r gives minimum A.

Chapter 8 Applications of Differentiation TMJC 2024 From 2 300h r , 3 300h r r  300 150 2         (iv) Let W be volume of liquid. d d d d d d H W H t t W  When A is a minimum, 3 150r  . 22 32 3 150 150W r H H H                   2 3d 150 d W H        2 3 d 1 100d 150 H t         2.4186 2.42 cm/s   The rate of increase of depth of liquid is 2.42 cm/s .  Use value of r found in (ii) When water is added to the can, depth increases but radius remains constant. 2 2 3 1 2 4 HW r H H H          refers to the volume of the can when A is minimum. When liquid is added to the empty can, the depth of the liquid increases, but the radius of the liquid surface 2W r H rremains as a constant, thus in , is constant and 2d d W rH  r and the value of is the value of r when A is 3 150r minimum (i.e ).

Chapter 8 Applications of Differentiation TMJC 2024 3 2015/MJC/Prelim/P2/Q2 Fig. 1 shows a rectangular piece of cardboard ABCD of sides 10 cm and 20 cm. A trapezium shape is cut out from each corner, to give the shape shown in Fig. 2. This shape consists of 2 isosceles triangles and 3 rectangles of different sizes. The remaining cardboard shown in Fig. 2 is folded along the dotte d lines, to form a closed triangular prism shown in Fig. 3. (i) Show that the volume 3 cmV of the closed triangular prism is given by   4 3 21 10 25 2505V h h h h    . [4] (ii) Use differentiation to find the maximum value of V, proving that it is a maximum. [5] Q3 Solution (i) 2 20 20 2y h y h     By Pythagoras Theorem,   22 2 2 2 2 2 5 25 10 25 10 h x x h x x x hx         Base of isosceles triangle 10 2 x  10 20 Fig. 1 Fig. 2 Fig. 3 h y A B D C h h x Note that if you used other variables in your solution, you are supposed to define your variables clearly.

Chapter 8 Applications of Differentiation TMJC 2024            2 2 3 2 4 3 2 1 10 22 1 2510 2 20 22 10 1 10 255 1 10 25 2505 1 10 25 250 (shown)5 V x hy h h h h h h h h h h h h h h                           (ii) Differentiate wrt x,   3 2d 1 4 30 50 250d 5 V h h hh     For maximum V, d 0.d V h    3 2 3 2 d 1 4 30 50 250 0d 5 4 30 50 250 0 V h h hh h h h           Using GC, 8.0902 cm (rejected 2 18.090 10), 3.0902 cm (rejected 0) or 2.5 cm. h x h h h         h 2.5 2.5 2.5 d d V h is maximum at 2.5.V h  Alternatively Using GC, 2 2 2.5 d 25.0 0d h V h    is maximum at 2.5.V h           4 3 2 3 1 maximum 2.5 10 2.5 25 2.5 250 2.55 70.3125 cm V      Express intermediate solutions in 5 significant figures. Obtain d d V h first before applying d 0d V h  for maximum value of V. Give reasons for rejection. Use GC to solve if possible 70.3125 is an exact value. No rounding off is allowed.